Question

two similar cones have volumes 12picu. units and 96pi cu. units. if the curved surface area of smaller cone is 15pi square units, what is curved surface area of larger one

Answer 1

since the cones are similar so l₁/l₂ = r₁/r₂
the the volumes are v₁ 12π unit³ , v₂ = 96π unit³ 
so curved surface area = 15π unit²

so v₂/v₁ =  96π/12π 
     (πr₁²l₁)/(πr₂²l₂) = 8             cancelling 3

⇒ (r₂²l₂)/(r₁²l₁) = 8         

⇒ $(\frac{r^2_2}{r^2_1})^2*\frac{l_2}{l_1}=8$

⇒ r₂³/r₁³ = 8   using l₁/l₂ = r₁/r₂

⇒ r₂/r₁ = 2

so curved surface area of the unknown cone be x

so cs₂/cs₁ = x/15π

⇒ πr₂l₂/πr₁l₁ = x/15π

⇒ r₂/r₁ x l₂/l₁ = x/15π

⇒ r₂²/r₁² = x/15π               using r₂³/r₁³ = 8

⇒ 4 = x/15π

⇒ x = 60π ANSWER

Answer 2

Let  R = radius of base and L = lateral height,  H = height of cone.

curved surface or lateral surface = A =  π R L ,  and    volume = V = π/3* R² H,

Similar cones have the same (cone) angle at the apex.  Hence the same ratios among R,  H and L.   Let Cone 1 be the larger cone.

       Let  R₂ = x R₁,  H₂ = x H₁  and  L₁ = x L₂

       V₁ / V₂ = π R₁² H₁ / π R₂² H₂  = x³ = 96 π / 12 π  = 8

                   Hence  x = 2 

     Lateral Surface areas A₁ / A₂ = π R₁ L₁ / π R₂ L₂   =  x²  = 4 ,    as x = 2

     A₁ = 4 A₂ = 4  * 15 π = 60 π square units.